일반 영역의 이중적분

Find the area of the region $D$ given in the previous exercise.

Find the area of the region $D$ given in the previous exercise.

Find the volume of the solid under the surface $z=2x+{y}^{2}$ and above the region bounded by $y={x}^{5}$ and $y=x.$

Find the volume of the solid under the plane $z=x-y$ and above the region bounded by $x=\tan \,y,x=-\tan \,y,$ and $x=1.$

Find the volume of the solid under the surface $z={x}^{3}$ and above the plane region bounded by $x=\sin \,y,x=-\sin \,y,$ and $x=1$

for values of $y$ between $y=\frac{-\pi }{2}\,\text{ and }\,y=\frac{\pi }{2}$

Find the volume of the solid situated in the first octant and determined by the planes $z=2,$ $z=0,x+y=1,x=0,\,\text{and}\,y=0.$

Find the volume of the solid situated in the first octant and bounded by the planes $x+2y=1,$ $x=0,y=0,z=4,\,\text{and}\,z=0.$

Find the volume of the solid bounded by the planes $x+y=1,x-y=1,x=0,z=0,$ and $z=10.$

Find the volume of the solid bounded by the planes $x+y=1,x-y=1,x+y=-1,$ $x-y=-1,z=1\,\text{and}\,z=0.$

Let ${S}_{1}$ and ${S}_{2}$ be the solids situated in the first octant under the planes $x+y+z=1$ and $x+y+2z=1,$ respectively, and let $S$ be the solid situated between ${S}_{1},{S}_{2},x=0,\,\text{and}\,y=0.$

a. Find the volume of the solid ${S}_{1}.$ b. Find the volume of the solid ${S}_{2}.$ c. Find the volume of the solid $S$ by subtracting the volumes of the solids ${S}_{1}\,\text{and}\,{S}_{2}.$

Let ${S}_{1}\,\text{and}\,{S}_{2}$ be the solids situated in the first octant under the planes $2x+2y+z=2$ and $x+y+z=1,$ respectively, and let $S$ be the solid situated between ${S}_{1},{S}_{2},x=0,\,\text{and}\,y=0.$

a. Find the volume of the solid ${S}_{1}.$ b. Find the volume of the solid ${S}_{2}.$ c. Find the volume of the solid $S$ by subtracting the volumes of the solids ${S}_{1}\,\text{and}\,{S}_{2}.$

Let ${S}_{1}\,\text{and}\,{S}_{2}$ be the solids situated in the first octant under the plane $x+y+z=2$ and under the sphere ${x}^{2}+{y}^{2}+{z}^{2}=4,$ respectively. If the volume of the solid ${S}_{2}$ is $\frac{4\pi }{3},$ determine the volume of the solid $S$ situated between ${S}_{1}$ and ${S}_{2}$ by subtracting the volumes of these solids.

Consider $X\,\text{and}\,Y$ two random variables of probability densities ${p}_{1}(x)$ and ${p}_{2}(y),$ respectively. The random variables $X\,\text{and}\,Y$ are said to be independent if their joint density function is given by $p(x,y)={p}_{1}(x){p}_{2}(y).$ At a drive-thru restaurant, customers spend, on average, $3$ minutes placing their orders and an additional $5$ minutes paying for and picking up their meals. Assume that placing the order and paying for/picking up the meal are two independent events $X$ and $Y.$ If the waiting times are modeled by the exponential probability densities

$\begin{array}{ccccccc}{p}_{1}(x)=\left\{\begin{array}{ll}\frac{1}{3}{e}^{-x/3} & x\ge 0, \\ 0 & \text{otherwise,}\end{array}\right. & & & \text{and} & & & {p}_{2}(y)=\left\{\begin{array}{ll}\frac{1}{5}{e}^{-y/5} & y\ge 0, \\ 0 & \text{otherwise,}\end{array}\right.\end{array}$

respectively, the probability that a customer will spend less than 6 minutes in the drive-thru line is given by $P[X+Y\le 6]=\iint_{D}p(x,y)dx\,dy,$ where $D=\{(x,y)\}|x\ge 0,y\ge 0,x+y\le 6\}.$ Find $P[X+Y\le 6]$ and interpret the result.

The integral $\int_{0}^{2\pi }\,\int_{0}^{1}\,\int_{r}^{1}dz\,dr\,d\theta$ represents the volume of a right cone.

The area of region enclosed by one petal of $r=\cos (4\theta ).$

The volume of the solid that lies between the paraboloid $z=2{x}^{2}+2{y}^{2}$ and the plane $z=8.$

The volume of the solid bounded by the cylinder ${x}^{2}+{y}^{2}=16$ and from $z=1$ and $z+x=2.$

The volume of the intersection between two spheres of radius 1, the top whose center is $(0,0,0.25)$ and the bottom, which is centered at $(0,0,0).$

In reality, it is very likely that the trash at the bottom of Mount Holly has become more compacted with all the weight of the above trash. Consider a density function with respect to height: the density at the top of the mountain is still density $400{\,\text{lb/ft}}^{3}$ and the density increases. Every $100$ feet deeper, the density doubles. What is the total weight of Mount Holly?